proyaop

Expanding: \(2x^2+16x+30<3x^2+19x+6\)

Combine like terms: \(x^2+3x-24>0\)

Quadratic formula: \(x={-3\pm\sqrt{105}\over2}\), which are the two roots r and s, used in the factoring of the quadratic (x - r)(x - s) = x^2 + 3x - 24.

To be positive, we take the extreme values in which both (x - r) and (x - s) are positive or both negative, so our solution interval is:

\((-\inf, {-3-\sqrt{105}\over2})\) U \(({-3 + \sqrt{105}\over2}, \inf)\)

What is the greatest integer n such that \(n^2 - 11n +24 \leq 17n - 8\)?

Combining like terms: \(n^2-28n+32\leq0\)

Using quadratic formula n = \({28\pm4\sqrt{41}\over2}=14\pm2\sqrt{41}\). Since our quadratic must be nonnegative, when factored (x - r)(x - s) where r and s are our roots, they must both be positive or both negative, so we take the extreme intervals: \((-\inf,14-2\sqrt{41}]\) U \([14 + 2\sqrt{41}, \inf)\)

To make a positive with 3 terms, you can have positive * positive * positive = positive, or positive * negative * negative = positive. To ensure x is the smallest, we probably take the latter option. To minimize x, (x - 5) and (x - 3) would be logical to be negative, whereas (x + 5) would be positive. If x + 5 > 0, subtract 5 from both sides and you get x > -5. The smallest integer greater than -5 is x = -4.

The function \(f\) satisfies \(f(\sqrt{2x} + x) = \frac{1}{x}\) for all \(x \ge \frac{1}{2}\), find \(f(4)\).

\(f(4) = f(\sqrt{2x}+x)={1\over{x}}\)

\(\sqrt{2x}+x=4\); \(\sqrt{2x} = -x + 4\); \(2x = x^2 - 8x + 16\); \(x^2-10x+16=0\); \((x-8)(x-2)=0\)

Note that x = 8 i s an extraneous solution, because \(\sqrt{2*8}\) does not equal to \(-8 + 4 = -4\), for a square root can't end up negative.

Thus, our only solution, x = 2. \(f(\sqrt{2(2)}+2)={1\over2}=f(4)\), so our final answer is 1/2.

\(4t^2\le-9t+12-14t+36\)

\(4t^2+23t-24\le0\)

\(t={-23\pm\sqrt{913}\over8}\) = r, s

Quadratic can be factored to (t - r)(t - s), where one of the binomials = 0, or one is negative and the other is positive to satisfy the inequality, thus the solution interval is: \([{-23 - \sqrt{913}\over8}, {-23 + \sqrt{913}\over8}]\)

Nice! I agree with your answer: personally, I prefer rationalizing the denominator: \({12\sqrt{2}\over\sqrt{5}}={12\sqrt{2}\over\sqrt{5}}*{\sqrt{5}\over\sqrt{5}}={12\sqrt{10}\over5}\). 

Combine like terms: \(x^3-x^2-7x>0\)

You can divide both sides by x, but consider the cases where x is positive and negative because the inequality sign will flip.

Case 1: x > 0:

\(x^2-x-7>0\)

Use quadratic formula to get the two "roots" \(x = {1\pm\sqrt{29}\over2}\). Since the quadratic has to be positive, then the factored form (x - r)(x - s) where r and s are roots have to have the same sign + + or both - - for all values of x. Thus, our union is \((-\inf, {1-\sqrt{29}\over2})\) U \(({1+\sqrt{29}\over2}, \inf)\). However, our domain restriction that x > 0 removes the left part of the union, so our answer for case 1 is \(({1 + \sqrt{29}\over2}, \inf)\).

Case 2: x < 0:

\(x^2-x-7<0\)

We get the same roots as above using the quadratic formula. However, since the quadratic has to be negative, then the factored form (x - r)(x - s) have to have opposite signs, + - or - +, as a result, we would have a solution of \(({1-\sqrt{29}\over2}, {1+\sqrt{29}\over2})\). However, we know that x has to be less than 0, so we restrict the solution interval to \(({1-\sqrt{29}\over2}, 0)\).

Hence, our solution interval for the entire problem is: \(({1-\sqrt{29}\over2},0)\) U \(({1+\sqrt{29}\over2},\inf)\)

Because the expression is a quadratic in t with a positive leading coefficient (so the parabola opens upwards), the minimal value of the quadratic would be at the vertex, a point that we can find using complete the square, aka vertex form:

\(t^2-9t-36+14t-60=t^2+5t-96\)

\(=1(t+{5\over2})^2-{25\over{4}}-96\)

\(=(t+2.5)^2-102.25\)

The vertex would be (-2.5, -102.25), so the real value of t = -2.25 would produce the smallest value of the quadratic.

You can also easily deduce this by observing that (t + 2.5)^2 will always be nonnegative for real values t, and because we want to minimize the expression, we want to also minimize (t + 2.5)^2, which will be 0, the smallest possible value, at t = -2.5.

A pretty simple problem using coordinate geometry:

Let ABCD be a square with side length 12, and let A be (0,0).

Line AM: y = x/2

Line DL: y = -2x + 12

The intersection of these two lines, P, is calculated by setting the two lines equal to each other:

x/2 = -2x + 12, 5x/2 = 12, x = 24/5

y = 24/10 = 12/5, so P has coordinates (24/5, 12/5)

Line OC: y = x/2 + 6

Line NB: y = -2x + 24

The intersection of these two lines, Q, is likewise calculated by setting OC = NB:

x/2 + 6 = -2x + 24, 5x/2 = 18, x = 36/5

y = 36/10 + 6= 48/5, so Q has coordinates (36/5, 48/5)

Using the distance formula between P and Q, PQ = sqrt[(36/5 - 24/5)^2 + (48/5 - 12/5)^2] = sqrt(144/25 + 1296/25) = sqrt(1440/25).

Simplifying, your final answer is \(12\sqrt{10}\over5\)

Okay, kudos if you caught it, but just because recreating the diagram and believing that PR is collinear is a fallacy that many fall into- the proper solution is below: